Ignore optparse if value is given

Question

I would like to have the ability to pass an argument w/o having to specify an option for optparse. If I do pass an option, it has to be a known option or else the script will fail.

Rsync the following file to destination

myscript.py filename

Rsync the following folder to destination (all this is figured out in function I create).

myscript.py -f foldername

The reason is, I have an array (or dict) that ties with the "foldername." If no options are passed, the argument given in the CLI is a file that's in the working folder where the user calls the script. Passing -f means to upload a folder whose value is stored in a dict (user can be in any directory, this folder's path is known ahead of time).

Am I better off adding options for both options? -f for file and -v for folder aka version?

Answer 1

I am always a fan of being explicit, but you can short circuit optparse.

# args[0] is always the executable's name
if len(args) > 1:
    parser.parse_args(args, options)
else:
    #open file

Answer 2

Treat -f as an option with no argument and use it to switch the behavior of the positional argument (i.e. foldername).

Answer 3

parser.parse_args() always returns two values:

• options, an object containing values for all of your options
• args, the list of positional arguments leftover after parsing options

(Totorial)

As such, you could do something like this:

parser = OptionParser()
parser.add_option("-f", "", action="store_true", dest="folder", default=False)
(options, args) = parser.parse_args(sys.argv)

fname = args[1] # name of folder or file

if options.folder:
    # do stuff for folder
else:
    # do stuff for file